A fair die is tossed until six is obtained on it. Let x be the number of required tosses, then the conditional probability P(x $\ge$ 5 | x > 2) is :
Solution
P(x $\ge$ 5 | x > 2) = ${{P(x \ge 5)} \over {P(x > 2)}}$<br><br>= $${{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^5}.{1 \over 6} + ....... + \infty } \over {{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6} + {{\left( {{5 \over 6}} \right)}^3}.{1 \over 6} + ...... + \infty }}$$<br><br>=$${{{{{{\left( {{5 \over 6}} \right)}^4}.{1 \over 5}} \over {1 - {5 \over 6}}}} \over {{{{{\left( {{5 \over 6}} \right)}^2}.{1 \over 6}} \over {1 - {5 \over 6}}}}} = {\left( {{5 \over 6}} \right)^2} = {{25} \over {36}}$$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
This question is part of PrepWiser's free JEE Main question bank. 143 more solved questions on Probability are available — start with the harder ones if your accuracy is >70%.